Question

the diameter of a road roller is 90 cm and 2m 10 cm length. If it takes 400 revolutions to level a playground, find the cost of levelling the ground at rs.2.25 per m^2

Answer 1

Given :

diameter of the roller = 90cm

length= 2m 10 cm = 2.1 m

so , radius of roller would be 45 cm or 0.45 m

The part which will touch the ground will be the C.S.A. of roller

now , C.S.A. of roller = 2πrh

= 2 × (22/7) × 0.45 × 2.1

= 5.94 m²

It takes 400 revolutions , so total area of ground = 5.94 × 400

= 2376 m²

now cost to level = rate × area

= 2376 × 2.25

= 5346

so ans = 5346 RS .

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Answer 2

☈ Given :-

Diameter of the roller = 90cm

length= 2m 10 cm = 2.1 m

So , radius would be 45 cm or 0.45 m

☈ To find :-

Find the cost of levelling the ground at ₹2.25 per square meter.

☈ Solution :-

The part which will touch the ground will be the C.S.A. of roller

now , C.S.A. of roller = 2πrh

➞ 2 × (22/7) × 0.45 × 2.1

➞ 5.94 m²

It takes 400 revolutions , area of ground

➞ 5.94 × 400

➞ 2376 m²

Now cost to level = rate × area

➞ 2376 × 2.25

➞ 5346

➞ 5346 Rs.

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